Chapter 13, Problem 13.201RP

The 2-Ib ball at A is suspended by an inextensible cord and given an initial horizontal velocity of v₀ , If $l=2$ ft, $v_0=0.3$ ft, and $y_B=0.4$ ft, determine the initial velocity v₀ so that the ball will enter the basket. (Hint: Use a computer to solve the resulting set of equations.)

To determine

The initial velocity $v_0$

Answer to Problem 13.201RP

Initial velocity $v_0$

$v_0=15.35ft/s$.

Explanation of Solution

Given information:

Weight of ball is $2lb$

$l=2ft$

$x_B=0.3ft$

$y_B=0.4ft$

The principle of conservation of energy is defined as,

When a particle moves under the action of conservation of forces, the sum of kinetic energy and potential energy of that particle remains constant.

$T_1+V_1=T_2+V_2$

Calculation:

At position 1,

$v_1=v_0$

At position 2,

The tension $Q$ is equal to zero

$\swarrow \sum F=0$

$Q+mgsin\theta =m{a}_n=m\frac{v_2^2}{l}$

Therefore,

$v_2=\sqrt{glsin\theta }$

According to conservation of energy,

$T_1+V_1=T_2+V_2$

Therefore,

$\begin{array}{l}\frac{1}{2}m{v}_0^2-mgl=\frac{1}{2}m{v}_2^2+mglsin\theta \\ v_0^2=v_2^2+2gl\left(1+sin\theta \right)\end{array}$

At position 2,

$\begin{array}{l}{x}_2= cos\theta \\ y_2= sin\theta \end{array}$

Assume $t_2$ be the time when the ball is in position 2,

For horizontal motion,

$\begin{array}{l}\dot{x}=-v_{2}sin\theta \\ x =x_2-(v_{2}sin\theta )(t-t_2)\end{array}$

At point B,

$\begin{array}{l}x=x_B \\ t=t_B\end{array}$

Therefore,

$(t_B-t_2)=\frac{lcos\theta -x_B}{v_{2}sin\theta }$

For vertical motion,

$\dot{y}=v_{2}cos\theta -g(t-t_2)$

$y=y_2+(v_{2}cos\theta )(t-t_2)-\frac{1}{2}g{(t-t_2)}^2$

At point B,

$y_B=l\sin \theta +(v_2\cos \theta )(t_B-t_2)-\frac{1}{2}g{(t_B-t_2)}^2$

According to given information,

$\begin{array}{l}{v}_2=\sqrt{64.4sin\theta }\\ (t_B-t_2)=\frac{2cos\theta -0.3}{v_{2}sin\theta }\\ y_B=2\sin \theta +(v_2\cos \theta )(t_B-t_2)-16.1{(}^{t_B}\end{array}$

For $\theta$

Try $\theta =30°$

$\begin{array}{l}{v}_2=\sqrt{64.4sin30°}=5.6745ft/s\\ (t_B-t_2)=\frac{2cos30°-0.3}{5.6745sin30°}=0.50473s\\ y_B=2\sin 30°+(5.6745\cos 30°)(0.50473)-16.1{(}^{0.50473}=-0.62116ft\end{array}$

Try $\theta =45°$

$\begin{array}{l}{v}_2=\sqrt{64.4sin45°}=6.7482ft/s\\ (t_B-t_2)=\frac{2cos45°-0.3}{6.7482sin30°}=0.23351s\\ y_B=2\sin 45°+(6.7482\cos 45°)(0.23351)-16.1{(}^{0.23351}=1.65060ft\end{array}$

Try $\theta =37.5°$

$\begin{array}{l}{v}_2=\sqrt{64.4sin37.5°}=6.2613ft/s\\ (t_B-t_2)=\frac{2cos45°-0.3}{6.2613sin30°}=0.33757s\\ y_B=2\sin 45°+(6.2613\cos 45°)(0.33757)-16.1{(}^{0.33757}=1.05972ft\end{array}$

Then,

Let, $u=\theta -30°$

Therefore,

$(u,y_B)=(0°,-0.62114ft),(7.5°,1.05972ft),(15°,1.65060ft)$

Therefore the quadratic curve equation will be,

$y_B=-0.62114+0.29678u-0.009688711u^2$

But, $y_B=0.4ft$

Therefore, above equation becomes,

$-0.009688711u^2+0.29678u-1.02114=0$

By solving we get,

$\begin{array}{ll}\begin{array}{l}u=3.95° \\ u=26.68°\end{array}\hfill & \hfill \end{array}$

Rejecting the second value, we get,

$\theta =30°+u=33.95°$

Now, try $\theta =33.95°$

$\begin{array}{l}{v}_2=\sqrt{64.4sin33.95°}=5.997ft/s\\ (t_B-t_2)=\frac{2cos33.95°-0.3}{5.997sin33.95°}=0.40578s\\ y_B=2\sin 33.95°+(5.997\cos 33.95°)(0.40578)-16.1{(}^{0.40578}=0.48462f\end{array}$

Therefore, new quadratic curve fits,

$(u,y_B)=(0°,-0.62114ft),(3.95°,0.48462ft),(7.5°,1.05972)$

Then the quadratic equation will be,

$y_B=-0.62114+0.342053907u-0.015725232u^2$

But, $y_B=0.4ft$

Therefore, above equation becomes,

$-0.015725232u^2+0.342053907u-1.02114=0$

Solve to find $u$

$u=3.572°$

Then,

$\theta =30°+3.572°=33.572°$

Now, try $\theta =33.572°$

$\begin{array}{l}{v}_2=\sqrt{64.4sin33.572°}=5.9676ft/s\\ (t_B-t_2)=\frac{2cos33.572°-0.3}{5.9676sin33.572°}=0.41406s\\ y_B=2\sin 33.572°+(5.9676\cos 33.572°)(0.41406)-16.1{(}^{0.41406}=0.40445ft\end{array}$

Above value of $y_B$ is close enough to $0.4ft$

Now substitute $\theta =33.572°$ And $v_2=5.9676ft/s$

Therefore,

$\begin{array}{l}{v}_0^2=v_2^2+2gl\left(1+sin\theta \right)\\ v_0^2={\left(5.9676ft/s\right)}^2+2\left(9.81m/s^2\right)\left(2\right)\left(1+sin33.572°\right)\\ v_0=15.35ft/s\end{array}$

Conclusion:

The initial velocity is equal to $v_0=15.35ft/s$